# What is i^i

$i^{i } = 0.207879576...$
$i = \sqrt{-1}$

If a is algebraic and b is algebraic but irrational then $a^b$ is transcendental. (Gelfond-Schneider Theorem)

Since i is algebraic but irrational, the theorem applies.

1. We know
$e^{ix}= \cos x + i \sin x$

Let $x = \pi/2$

2. $e^{i \pi/2} = \cos \pi/2 + i \sin \pi/2$

$\cos \pi/2 = \cos 90^\circ = 0$

$\sin 90^\circ = 1$
$i \sin 90^\circ = (i)*(1) = i$

3. Therefore
$e^{i\pi/2} = i$
4. Take the ith power of both sides, the right side being $i^i$ and the left side =
$(e^{i\pi/2})^{i}= e^{-\pi/2}$
5. Therefore
$i^{i} = e^{-\pi/2} = .207879576...$